Solve {rrr}{3}&{-4}&{6}{3}&{0}&{1}{4}&{2}&{1}

$$\left| \begin{array} { r r r } { 3 } & { - 4 } & { 6 } \\ { 3 } & { 0 } & { 1 } \\ { 4 } & { 2 } & { 1 } \end{array} \right|$$

Evaluate

$$26$$

Steps Using Sarrus' Rule

Find the determinant of the matrix using the method of diagonals.

$$det(\left(\begin{matrix}3&-4&6\\3&0&1\\4&2&1\end{matrix}\right))$$

Extend the original matrix by repeating the first two columns as the fourth and fifth columns.

$$\left(\begin{matrix}3&-4&6&3&-4\\3&0&1&3&0\\4&2&1&4&2\end{matrix}\right)$$

Starting at the upper left entry, multiply down along the diagonals, and add the resulting products.

$$-4\times 4+6\times 3\times 2=20$$

Starting at the lower left entry, multiply up along the diagonals, and add the resulting products.

$$2\times 3+3\left(-4\right)=-6$$

Subtract the sum of the upward diagonal products from the sum of the downward diagonal products.

$$20-\left(-6\right)$$

Subtract -6 from 20.

$$26$$

Steps Using Expansion by Minors

Find the determinant of the matrix using the method of expansion by minors (also known as expansion by cofactors).

$$det(\left(\begin{matrix}3&-4&6\\3&0&1\\4&2&1\end{matrix}\right))$$

To expand by minors, multiply each element of the first row by its minor, which is the determinant of the 2\times 2 matrix created by deleting the row and column containing that element, then multiply by the element's position sign.

$$3det(\left(\begin{matrix}0&1\\2&1\end{matrix}\right))-\left(-4det(\left(\begin{matrix}3&1\\4&1\end{matrix}\right))\right)+6det(\left(\begin{matrix}3&0\\4&2\end{matrix}\right))$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the determinant is ad-bc.

$$3\left(-2\right)-\left(-4\left(3-4\right)\right)+6\times 3\times 2$$

Simplify.

$$3\left(-2\right)-\left(-4\left(-1\right)\right)+6\times 6$$

Add the terms to obtain the final result.

$$26$$

Factor

$$2\times 13$$